What is #f(x) = int x/sqrt(x-2) dx# if #f(3)=-1 #?
1 Answer
Mar 22, 2018
Explanation:
Let
#f(x) = int x/sqrt(u)dx#
#f(x) = int (u +2)/sqrt(u) du#
#f(x) = int u/sqrt(u) + 2/sqrt(u) du#
#f(x) = int u^(1/2) + 2u^(-1/2) du#
#f(x) = 2/3u^(3/2) + 4u^(1/2) + C#
#f(x) = 2/3(x - 2)^(3/2) +4(x- 2)^(1/2) + C#
Now we solve for
#-1 = 2/3(3- 2)^(3/2) + 4(3 -2)^(1/2) + C#
#-1 = 2/3 + 4 + C#
#-17/3 = C#
Therefore the function is
#f(x) = 2/3(x -2)^(3/2) + 4(x- 2)^(1/2) - 17/3#
Hopefully this helps!